Question

The stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of $2V$ volts. If e and m are the charge and the mass of electron respectively, then the value of $\frac{h}{\lambda }$ is:
(where $\lambda$ is the de Broglie wavelength of the electron)

• A. $\sqrt{meV}$
• B. $\sqrt{2meV}$
• C. $2\sqrt{meV}$
• D. $2meV$

Answer 1

The correct option is C $2\sqrt{meV}$

Given:
Charge of electron = e
Potential difference applied = $2V$
Mass of electron = m

Let $v$ be the veloctiy of the electron. Kinetic energy would be =$\frac{1}{2}m{v}^2$

Kinetic energy gained by electron due to the applied potential
$=e\left(2V\right)$
$\therefore 2eV=\frac{1}{2}m{v}^2⟹2meV=\frac{1}{2}{m}^2{v}^2⟹4meV=(mv{)}^2⟹2\sqrt{meV}=mv$

de Broglie wavelength of the electron given by:
$\lambda =\frac{h}{mv}⟹\frac{h}{\lambda }=mv$
Since
$mv=2\sqrt{meV}$
$\frac{h}{\lambda }=2\sqrt{meV}$